Let $x, y, z$ be positive real numbers such that $x+y+z=1$. Prove that:
$$\sum_{cyc}\frac{x^3}{\left ( x+y \right )^2}\geq \frac{1}{4}$$
Solution
It is pretty well known that:
$$\frac{x_1 ^m}{a_1 ^{m-1}}+\frac{x_2 ^m}{a_2 ^{m-1}}+...+\frac{x_n ^m}{a_n ^{m-1}}\geq \frac{(x_1 +x_2 + ... +x_n)^m}{(a_1 +a_2 + ... +a_n )^{m-1}}$$
where $m$ is a natural number greater or equal to $2$. Hence:
$$ \sum_{cyc} \frac{x^3 }{(x+y)^2} \geq \frac{(x+y+z)^3}{(x+y+y+z+z+x)^2}=\frac{1}{(2x+2y+2z)^2} =\frac{1}{4(x+y+z)^2}=\frac{1}{4}$$
completing the exercise.
The exercise can also be found in mathematica.gr
$$\sum_{cyc}\frac{x^3}{\left ( x+y \right )^2}\geq \frac{1}{4}$$
Solution
It is pretty well known that:
$$\frac{x_1 ^m}{a_1 ^{m-1}}+\frac{x_2 ^m}{a_2 ^{m-1}}+...+\frac{x_n ^m}{a_n ^{m-1}}\geq \frac{(x_1 +x_2 + ... +x_n)^m}{(a_1 +a_2 + ... +a_n )^{m-1}}$$
where $m$ is a natural number greater or equal to $2$. Hence:
$$ \sum_{cyc} \frac{x^3 }{(x+y)^2} \geq \frac{(x+y+z)^3}{(x+y+y+z+z+x)^2}=\frac{1}{(2x+2y+2z)^2} =\frac{1}{4(x+y+z)^2}=\frac{1}{4}$$
completing the exercise.
The exercise can also be found in mathematica.gr
Here is a solution I found.
ReplyDeleteUsing the Cauchy-Schwarz and AM-GM Inequalities we have
$$\begin{aligned}
\sum\frac{x^3}{(x+y)^2}\geq\sum\frac{x^3}{2(x^2+y^2)}&=\sum\left(\frac{x}{2}-\frac{xy^2}{2(x^2+y^2)}\right)\\&\geq\sum\left(\frac{x}{2}-\frac{y}{4}\right)\\&=\frac{x+y+z}{4}\\&=\frac{1}{4}
\end{aligned}$$
and the proof is completed.